Why there isn't any exact formula to calculate summation of n terms of a HP?

Question

When I search over the net to find such a formula all I find is the approximations but not exact. So, is it even possible to find the exact formula or is it the case that we will never be able to find an exact formula for this?

Answer 1

The only form (other than the original) that doesn't use special functions:

$$\sum_{k=1}^n\frac{r^k}k=\int_0^r\frac{x^n-1}{x-1}dx$$

You are mainly interested in $r=1$.

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$$\sum_{k=1}^n\frac1k=\gamma+\psi^{(0)}(n)$$

where $\gamma:=\lim\limits_{n\to\infty}\sum_{k=1}^n\frac1k-\int_1^n\frac1xdx$ is the Euler-Mascheroni constant and $\psi^{(0)}(n):=\frac{d}{dx}\ln\left(\Gamma(n)\right)$ is the digamma function.

Also note that...

$$\sum_{k=a}^b\frac1{dk+c}=\frac{\sum_{k=1}^{b+\frac cd}\frac1k-\sum_{k=1}^{a+\frac cd-1}\frac1k}d$$

and then use the above.

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If you are interested in $\sum_{k=1}^n\frac1{k^p}$, it can be written as

$$\sum_{k=1}^n\frac1{k^p}=\zeta(p)-\frac{\psi^{(p-1)}(n+1)}{(-1)^p(p-1)!}\tag{$p\in\mathbb N$}$$

where $\psi^{(p)}(n)$ is the polygamma function and $\zeta(p)$ is the Riemann zeta function.

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If you allow the Hurwitz zeta function $\zeta(p,n)$,

$$\sum_{k=1}^n\frac1{k^p}=\zeta(p)-\zeta(p,n-1)$$